Question

find the range of the function f(x) = definite integral -4 to x sqrt(16 - t^2) dt.

Answer 1

[-4, 4]
[-4, 0]
[0,-4]
[0,4pi]
[0, 8pi]

Answer 2

\[I=\int\limits_{-4}^{x}\sqrt{16-t^2} dt\]
for range
\[\sqrt{16-t^2}\ge 0,16-t^2\ge 0,16\ge t^2\]
t^2 <=16
\[\left| t \right|<=4 \]
\[\left[ -4,4 \right]\]

Answer 3

are you sure that isn't the domain?

Answer 4

no ,it is domain.

Answer 5

\[I=\int\limits_{-4}^{x}\sqrt{16-t^2}*1~dt=[ \sqrt{16-t^2}* t]({-4 \rightarrow x})-\int\limits_{-4}^{x}\frac{ -2t }{2\sqrt{16-t^2} }*t*dt\]
\[=\sqrt{16-x^2}-0-\int\limits_{-4}^{x}\frac{ 16-t^2-16 }{\sqrt{16-t^2} }dt\]
\[=\sqrt{16-x^2}-\int\limits_{-4}^{x}\sqrt{16-t^2}dt+16\int\limits_{-4}^{x}\frac{ dt }{ \sqrt{4^2-t^2} }\]
\[I+I=\sqrt{16-x^2}+16\sin^{-1} \frac{ t }{ 4 }\left({-4 \rightarrow x} \right)\]
\[I=\frac{ \sqrt{16-x^2} }{ 2 }+8\left( \sin^{-1} \frac{ x }{4 }-\sin^{-1} \frac{ -4 }{4 } \right)\]

Answer 6

i have to go.

Answer 7

Think about what the integrand represents: \(\sqrt{16-t^2}\). You know \([-4,4]\) is the domain of \(f(x)\). Since \(t\) is just a placeholder, let's call it \(x\) for a moment. Plotting this, you have